Question

If $b^2<2ac$ and $a,b,c,d\in R,$ then the number of real roots of the equation $a{x}^3+b{x}^2+cx+d=0$ are

• A. 1.00
• B. 1.0
• C. 01
• D. 1
• E. 01.0
• F. 001
• G. 01.00

Answer 1

Answer: (A), (B), (C), (D), (E), (F), (G)

Given: $a{x}^3+b{x}^2+cx+d=0,$ $b^2<2ac$ and $a,b,c,d\in R$

Let $\alpha ,\beta ,\gamma$ be the roots of $a{x}^3+b{x}^2+cx+d=0$

$\therefore \text{sum of roots}=\alpha +\beta +\gamma =-\frac{b}{a}$
$\text{sum of roots taken two at a time}=\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}$
$\text{product of roots}=\alpha .\beta .\gamma =\frac{-d}{a}$

Now, ${\alpha }^2+{\beta }^2+{\gamma }^2=(\alpha +\beta +\gamma {)}^2-2(\alpha \beta +\beta \gamma +\gamma \alpha )$

$\Rightarrow {\alpha }^2+{\beta }^2+{\gamma }^2=(-\frac{b}{a}{)}^2-2(\frac{c}{a})$
$\Rightarrow {\alpha }^2+{\beta }^2+{\gamma }^2=\frac{b^2-2ac}{a^2}$
We have given that $b^2-2ac$ is negative,
$\therefore {\alpha }^2+{\beta }^2+{\gamma }^2<0,$ which is not possible if all $\alpha ,\beta ,\gamma$ are real. Therefore atleast one roots is non-real, but complex roots occurs in pair. Hence given cubic equation has one real root and two non real roots.